Proportion of non-commuting couples in a non-abelian group Consider $G$ a non-abelian group, and let $$n(G)=\frac{1}{|G|^2}\#\{(x,y)\in G^2/xy=yx\}$$
I am trying to prove that $n(G)\leq\frac{5}{8}$, that is that at least 3/8ths of the multiplication table are non-symmetrical elements (entries $x,y$ and $y,x$ in the multiplication table are distinct).

What I've tried doing so far is picking $a,b$ that do not commute, and looking at the multiplication table for elements that do or do not commute.
The first line and column consist of the unit element which commutes with all elements, and the diagonal is trivially commuting couples. That means that there are at least $3|G|-2$ couples that do commute in the table.
As for non-commuting elements, I've only come up with $(a,b),(a^{-1},b),(a^{-1},b^{-1}),(a,b^{-1})$ but have absolutely no means of proving they're distinct.
I've also tried looking at elements that commute with $a$ and prove they don't commute with $b$, but it does not lead me anywhere...
 A: We have, using Lagrange and proper subgroups:
$$\begin{align*}
  n(G) &= \frac{1}{|G|^2}\sum_{x\in G}|C_G(x)|\\&=\frac{1}{|G|^2}\big(\sum_{x\in Z_G}|C_G(x)|+\sum_{x\notin Z_G}|C_G(x)|\big)\\&\leq\frac{1}{|G|^2}\big(\sum_{x\in Z_G}|G|+\sum_{x\notin Z_G}\sqrt{|G|}\big)\\&\leq\frac{1}{|G|^2}\big(|Z_G||G|+(|G|-|Z_G|)\sqrt{|G|}\big)\\&\leq\frac{1}{|G|^2}\big(\sqrt{|G|}|G|+(|G|-|Z_G|)\sqrt{|G|}\big)\\&\leq\frac{(2|G|-|Z_G|)}{|G|^2}\sqrt{|G|}\end{align*}$$
We then distinguish several cases, given that for $G$ to be non-abelian, we must have $|G|\geq6$ (since the two products of two non-commuting elements $x,y$ must be different and not in $\{e,x,y\}$, and $x²,y²\notin\{x,y,xy,yx\}$ but cannot both be the identity).
In fact, the order of a non-abelian subgroup smaller than 10 is in 6,8,10 (see here). Any group bigger than 10 verifies the inequality with a trivial center (study the function $x\mapsto (2x-1)x^{-3/2}$).
}
We are therefore left with the cases 6 and 8 on our hands.
n=6
The dihedral group of order 6 is centerless, however we do have 18 non-symmetric ordrered couples in its Cayley table.
n=8


*

*The Dicyclic group has two elements in its center and therefore verifies the inequality.

*The Dihedral group of order 8 verifies the inequality is an equality by painstakingly checking its Cayley table (or simply noticing its center has two elements too).

Please note this was a very 'manual' job, I think there's a faster way of proving it...
We have also proven it's the best possible inequality we can get for $n(G)$ since the equality is reached.
